# Prove $(\frac{q(q+1)}{p}) = (\frac{1+q^{-1}}{p})$ for p>2 a prime, and any $q \in \mathbb{Z^+}$

For p>2 a prime, and any $q \in \mathbb{Z^+}$,

show that $(\frac{q(q+1)}{p}) = (\frac{1+q^{-1}}{p})$ where the terms are legendre terms.

I saw this result as part of a proof, and it was simply assumed, like the result was trivial/obvious.. but I cannot seem to see why this is so.. is it a typo? or can someone please explain/prove why?

Thanks heaps!

-
$\left(\dfrac {q^2} p\right) = 1$ for all $q \in \Bbb Z^+$. Care needs to be taken that $q \ne 0 \pmod p$ for the RHS to make sense. –  Lord_Farin Oct 25 '12 at 10:58
$q(q+1)=q^2(1+q^{-1})$. $(ab|p)=(a|p)(b|p)$. –  Gerry Myerson Oct 25 '12 at 12:09
if we had GCD(q,p) = 1, does it necessarily follow that $GCD(1+q^{-1},p)=1$ also? since this would be required to use the rule: (ab|p)=(a|p)(b|p). –  Jake Fresco Oct 25 '12 at 18:51
Jake, if you want to be sure I see a comment addressed to me, you have to write @Gerry. Anyway, if, say, $p$ divides $a$, then multiplicativity still holds, as both sides are zero. –  Gerry Myerson Oct 25 '12 at 23:05